Math 54 Fall 2017 Practice Final Exam Exam date: 12/14/17 Time Limit: 170 Minutes Name: Student ID: GSI or Section: This exam contains 9 pages (including this cover page) and 10 problems. Problems are printed on both sides of the pages. Enter all requested information on the top of this page. This is a closed book exam. No notes or calculators are permitted. We will drop your lowest scoring question for you. You are required to show your work on each problem on this exam. Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations will receive partial credit. If you need more space, there are blank pages at the end of the exam. Clearly indicate when you have used these extra pages for solving a problem. However, it will be greatly appreciated by the GSIs when problems are answered in the space provided, and your final answer must be written in that space. Please do not tear out any pages. Do not write in the table to the right. Problem Points Score 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total: 100
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1. (10 points) Consider the matrix 1 1 2 A x = x 2 3 0 1 1 (a) (5 points) Find all values of x such A x is invertible. Solution: We compute the determinant. 1 1 2 x 2 3 0 1 1 = 1 2 x 3 + 1 1 x 2 = x 1 Therefore, A x is invertible as long as x 1. (b) (5 points) Compute A 1 2. Solution: We proceed by row reduction. 1 1 2 1 0 0 1 1 2 1 0 0 1 0 1 1 0 1 2 2 3 0 1 0 0 0 1 2 1 0 0 1 1 0 0 1 0 1 1 0 0 1 0 1 1 0 0 1 0 0 1 2 1 0 Therefore, 1 0 0 1 1 1 0 1 0 2 1 1 0 0 1 2 1 0 1 1 1 A 1 2 = 2 1 1. 2 1 0
2. (10 points) Consider the matrices 5 1 2 5 1 2 A = 1 5 2 B = 1 5 2 2 2 2 2 1 2 (a) (2 points) Which of these matrices is orthogonally diagonalizable? Solution: A is orthogonally diagonalizable since A is symmetric. B is not orthogonally diagonalizable since B is not symmetric. (b) (2 points) Find the eigenvalues of the orthogonally diagonalizable matrix from part (a) Solution: We have 5 λ 1 2 det(a λi) = 1 5 λ 2 2 2 2 λ = (5 λ) 5 λ 2 2 2 λ + 1 2 2 2 λ + 2 1 5 λ 2 2 so the eigenvalues are 0 and 6. = (5 λ)(λ 2 7λ + 6) + λ 6 + 24 + 4λ = λ 3 + 12λ 2 36λ λ(λ 2 12λ + 36) = λ(λ 6) 2
(c) (6 points) Find an orthogonal matrix P consisting of eigenvectors of the orthogonally diagonalizable matrix from part (a). Solution: We start by finding an orthogonal basis of each eigenspace. For E 0, we find the null space of A. 5 1 2 1 5 2 1 0 1/2 1 5 2 0 24 12 0 2 1 2 2 2 0 12 6 0 0 0 1 so 1 is an orthogonal basis for E 0. For E 6, we find the null space of A 6I 2 1 1 2 1 1 2 1 1 2 0 0 0 2 2 4 0 0 0 1 2 Thus, 1, 0 is a basis for E 6. Applying the Gram-Schmidt algorithm gives 0 1 1 1 that 1, 1 is an orthogonal basis for E 6. Normalizing, we get 0 1 1/ 6 1/ 2 1/ 3 P = 1/ 6 1/ 2 1/ 3 2/ 6 0 1/ 3
3. (10 points) Label the following statements as True or False. The correct answer is worth 1 point and a brief justification is worth 1 point. Credit for the justification can only be earned in conjunction with a correct answer. No points will be awarded if it is not clear whether you intended to mark the statement as True or False. (a) (2 points) If σ is the largest singular value of an m n matrix A, then Av σ v for all v R n where u = u u. Solution: True. By the singular value decomposition, there is an orthonormal basis {v 1,..., v n } of R n and an orthonormal basis {u 1,..., u m } of R m such that Av i = σ i u i and σ i σ are the singular values of A. Then, For any v = c 1 v 1 +... + c n v n, we have as desired. Av Av = c 2 1σ 2 1 +... + c 2 nσ 2 n σ 2 (c 2 1 +... + c 2 n) = σ 2 v v (b) (2 points) If A and B are similar n n matrices so that A = P BP 1, then P y(t) is a solution to x (t) = Ax(t) for any y(t) such that y (t) = By(t). Solution: True. We have y (t) = By(t) = P 1 AP y(t). After applying P to both sides, we get P y (t) = AP y(t) as we needed since d dt P y(t) = P y (t). (c) (2 points) Only square matrices can be squared. Solution: True. In order for A 2 to make sense for an m n matrix, we must have m = n. (d) (2 points) If, is an inner product on R n such that v, v = v v for all v R n then u, w = u w for all u, w R n. Solution: True. By properties of inner products, we have u, w = 1 2 ( u + w, u + w u, u w, w ) = ((u + w) (u + w) u u w w) = u w. 1 2 (e) (2 points) The set of solutions to y + by + cy + dy = 0 is always a vector space of dimension three. Solution: True. The solutions of a homogeneous equation always form a vector space so we need to check the dimension. The solutions of this equation can recast as solutions to an equation of the form x (t) = Ax(t) where A is 3 3 matrix by setting x 1 = y, x 2 = y, and x 3 = y. Such an equation always has a 3-dimensional solution space spanned by the columns of a fundamental matrix (such as e At ).
4. (10 points) Consider the function on R 3 given by for x, y R 3. x y = 3 mx m y m m=1 (a) (3 points) Show that this is an inner product on R 3. Solution: We need to check four properties: x y = y x, (x + v) y = x y + v y, (cx) y = c(x y), and x x 0 and only zero when x = 0 for x, y, v R 3 and c R. For the first, we have For the second, (x + v) y = For the third, For the last x y = 3 mx m y m = m=1 3 m(x m + v m )y m = m=1 (cx) y = 3 my m x m = y x. m=1 3 mx m y m + m=1 3 mv m y m = x y + v y. m=1 ( 3 3 ) mcx m y m = c mx m y m = c(x y). m=1 x x = m=1 3 mx 2 m 0 i=1 since each term is positive and equal to zero only when x 2 1 = x2 2 = x2 3 = 0, i.e., x = 0. We have verified that this is an inner product.
1 0 (b) (4 points) Find an orthogonal basis for V = Span 1, 1 1 1 inner product. with respect to the Solution: We apply the Gram-Schmidt procedure. We have 1 u 1 = 1 1 and 1 0 1 1 0 1 1 1 0 u 2 = 1 1 = 1 1 1 1 1 6 1 1 1 1 1 1 form an orthogonal basis of V (c) (3 points) Find the closest vector in V (from (b)) to x = computed with the inner product. 1 1 = 1 6 1 1 7 5 0 0 when the distance is 1 Solution: The closest vector is the orthogonal projection. To simplify computations, we replace u 1, u 2 from part (b) with the orthogonal basis w 1 = u 1, w 2 = 6u 2. We have Proj V (x) = x w 1 w 1 + x w 2 w 2 = 1 w 1 w 1 w 2 w 2 2 w 1 5 58 w 2 = 1 24 6. 58 54
5. (10 points) Suppose that U is an orthogonal n n matrix that is orthogonally diagonalizable. (a) (4 points) Show that the only eigenvalues of U are ±1. Solution: Since U is orthogonal, U T = U 1. Since U is orthogonal diagonalizable, U is symmetric U T = U. Thus, we have U 2 = I. If v is an eigenvector with eigenvalue λ, we have λ 2 v = U 2 v = v so λ 2 = 1. That is, λ = ±1 as desired. (b) (3 points) Show that if U has only 1 as an eigenvalue then U is the n n identity matrix. Solution: In that case, we can diagonalize U to get since the only eigenvalue of U is 1. U = P IP 1 = P P 1 = I (c) (3 points) Give an example of an orthogonal and orthogonally diagonalizable matrix U with entries other than ±1. Solution: One simple example is U = 1 [ ] 1 1. 2 1 1
6. (10 points) Suppose that y 1 and y 2 are solutions to y + b(t)y + c(t)y = 0 and W (t) = W [y 1, y 2 ](t) is their Wronskian. (a) (4 points) Using only the defintion of W (t), show that W (t) satisfies the equation W (t) = b(t)w (t). Deduce from your calculation that if W (0) 0 then W (t) 0 for all t. Solution: We have that W (t) = y 1(t) y 1 (t) y 2 (t) = y 1(t)y 2(t) y 2 (t)y 1(t). y 2 (t) Differentiating, we get W (t) = y 1(t)y 2(t) + y 1 (t)y 2(t) y 2(t)y 1(t) y 2 (t)y 1(t) = y 1 (t) [ b(t)y 2(t) c(t)y 2 (t) ] y 2 (t) [ b(t)y 1(t) c(t)y 1 (t) ] as desired. This computation shows that is never zero if W (0) = C 0. (b) (2 points) Show that if y 1 (t) 0 then = b(t)(y 1(t)y 2 (t) y 2(t)y 1 (t) = b(t)w (t) W (t) = Ce t 0 b(s) ds ( y2 y 1 ) = W (y 1 ) 2 Solution: We compute using the product rule. ( y2 y 1 ) = y 2 y 1 y 1 y 2 (y 1 ) 2 = W (y 1 ) 2
(c) (4 points) Given that y 1 (t) = e t2 is a solution to y + 4ty + (4t 2 + 2)y = 0, find another linearly independent solution using parts (a) and (b). Solution: By part (a), W (t) = Ce t 0 b(s) ds = Ce 2t2. By rescaling the unknown linearly independent solution, we can assume that C = 1. By part (b), we get that ( ) y2 = e 2t2 = 1. e 2t2 Thus, we can take y 1 y 2 (t) = ty 1 (t) = te t2.
7. (10 points) Consider the differential equation y 6y + 14y 14y + 5y = 0. (a) (6 points) Find the general solution of the above differential equation. (Hint: r = 2 + i is a root of the characteristic polynomial.) Solution: By the standard substitution x 1 = y, x 2 = y, x 3 = y, and x 4 = y, we can translate this to an equation of the form x (t) = Ax(t) where A is 4 4 matrix with characteristic polynomial is r 4 r 3 + 14r 2 14r + 5 = 0. We know that 2 + i is a root so 2 i must also be a root since we are working with a real polynomial. Therefore, (r (2 + i))(r (2 i)) = r 2 4r + 5 divides the polynomial. Doing long division gives that r 4 r 3 + 14r 2 14r + 5 = (r 2 4r + 5)(r 2 2r + 1) so the only other root is the double root r = 1. Thus, the general solution is y(t) = C 1 e 2t cos(t) + C 2 e 2t sin(t) + C 3 e t + C 4 te t. (b) (4 points) Solve the initial value problem where y(0) = 1, y (0) = 3, y (0) = 5, and y (0) = 5. Solution: From the general solution in the first part, we calculate y (t) = (2C 1 + C 2 )e 2t cos(t) + ( C 1 + 2C 2 )e 2t sin(t) + (C 3 + C 4 )e t + C 4 te t, y (t) = (3C 1 + 4C 2 )e 2t cos(t) + ( 4C 1 + 3C 2 )e 2t sin(t) + (C 3 + 2C 4 )e t + C 4 te t, y (t) = (2C 1 + 11C 2 )e 2t cos(t) + ( 11C 1 + 2C 2 )e 2t sin(t) + (C 3 + 3C 4 )e t + C 4 te t. By putting t = 0, we get a system of linear equations corresponding to an augmented matrix which we row reduce below. 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 2 1 1 1 3 3 4 1 2 5 0 1 1 1 1 0 4 2 2 2 0 1 1 1 1 0 0 2 2 2 2 11 1 3 5 0 11 2 3 3 0 0 10 8 8 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 0 0 2 2 2 0 1 0 0 0 0 0 1 0 0. 0 0 0 2 2 0 0 0 1 1 So the solution to the IVP is y(t) = e 2t cos(t) + te t.
8. (a) (4 points) Find a basis of solutions to the system [ ] x 2 1 (t) = x(t) 0 1 [ ] 2 1 Solution: Since A = is upper triangular, its eigenvalues are the entries on 0 1 the diagonal. We compute the eigenspaces. [ ] [ ] 1 1 1 E 1 = Nul(A Id) = Nul = span 0 0 1 [ ] [ ] 0 1 1 E 2 = Nul(A 2Id) = Nul = span 0 1 0 Therefore a basis of solutions to the given system is { [ ] [ ]} e t 1, e 2t 1 1 0 (b) (6 points) Find a particular solution to [ ] [ x 2 1 t (t) = x(t) + 1 e 2t ] 0 1 0 Solution: Variation of parameters tells us that we can find a solution of the form X(t)v(t) where [ e t e X(t) = 2t ] e t 0 and v(t) is a vector satisfying We therefore find that v (t) = us the solution [ X(t)v t (t) = 1 e 2t ] 0 [ ] [ ] 0 0 t 1, and so we can take v(t) =. This gives log t [ e X(t)v(t) = 2t ] log t 0
9. (10 points) Consider the heat equation u t = u 2 2 x 2. (a) (6 points) Find the solution defined for 0 x π and t 0 that satisfies u(0, t) = u(π, t) = 0 and u(x, 0) = sin(3x) 5 sin(5x). Solution: The general solution with L = π and β = 2 and Dirichlet boundary condition is given by c n sin(nx)e 2n2t. n=1 For u(x, 0) = sin(3x) 5 sin(5x), we must have that the c n are the coefficients of the Fourier sine series of this function. However, sin(3x) 5 sin(5x) is already a sine series so we get c 3 = 1, c 5 = 5 and all the other coefficients vanish. Therefore, u(x, t) = sin(3x)e 18t 5 sin(5x)e 50t. (b) (4 points) Suppose that v(x, t) is the solution defined for 0 x π and t 0 that satisfies v(0, t) = 0, v(π, t) = 1 and v(x, 0) = sin(3x) 5 sin(5x). Find lim t v(x, t). Solution: We can write v(x, t) = w(x, t) + x π where w(x, t) is a solution to the heat equation satisfying the Dirichlet boundary conditions and w(x, 0) = sin(3x) 5 sin(5x) x π. Thus, we see that v(x, t) will be of the form v(x, t) = x π + b n sin(nx)e 2n2 t n=1 where the b n are the coefficients of the Fourier sine series for w(x, 0). Therefore, lim v(x, t) = x t π + lim t b n sin(nx)e 2n2t = x π + 0 = x π. n=1
10. (10 points) Consider the function ζ(s) = n=1 1 n s. Euler famously computed ζ(2) = π 2 /6 in 1738. Evaluate ζ(4) by computing the Fourier series of f : [ π, π] R, given by f(x) = π 4 x 4. You may assume Euler s result. Solution: Since f is even, b n = 0 for all n. We now compute the other coefficients. We have π a 0 = (π 4 x 4 ) dx = π 4 x x5 π 5 = 2π 5 2π5 5 = 8π5 5 π and for n 1, π π π (π 4 x 4 ) cos nx dx = π 4 cos nx dx x 4 cos nx dx π π π [ x 4 π sin nx = n 4 π ] x 3 sin nx dx π n π = 4 [ x3 π cos nx n n + 3 π ] x 2 cos nx dx π n π = 4 [ ( π 3 cos nπ ( π) 3 ) cos( nπ) + 3 π n n n π = 4 [ ( ( 1) n 2π 3 ) + 3 π ] x 2 cos nx dx n n n π = ( 1)n+1 8π 3 n 2 + 12 [ x 2 π sin nx n 2 n 2 π ] x sin nx dx π n π = ( 1)n+1 8π 3 n 2 24 [ x cos nx π n 3 n + 1 π ] cos nx dx π n π [( )] π cos nπ ( π) cos( nπ) Thus, = ( 1)n+1 8π 3 n 2 + 24 n 3 = ( 1) n+1 π 8(n2 π 2 6) n 4. π 4π 4 5 8 ( 1) n n2 π 2 6 n 4 cos nx n=1 n ] x 2 cos nx dx is the Fourier series of f, and because the 2π-periodic extension of f is piecewise differentiable and continuous at π with f(π) = 0, we have n=1 0 = f(π) = 4π4 5 8π2 n=1 Rearranging the above, it follows that ( 1 ζ(4) = n 4 = 1 ) 8π 2 1 48 n 2 4π4 5 n=1 1 n 2 + 48 1 n 4. = 1 48 n=1 ) (8π 2 ζ(2) 4π4 = π4 5 90.
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